Given a dictionary of words and a string with all spaces removed, determine whether the string is composed of valid words

Problem Statement

Given a dictionary of words and a string with all spaces removed, determine whether the string is composed of valid words

Examples

helloworld-> hello world (valid)

isitniceinhere-> is it nice in here (valid)

zxyy-> invalid

Time Complexity Requirement

O(n^2)

Algorithm

True/False array T of size equal to the input. T[i] = true => the substring starting from i is composed of valid words.

Assume API method IsValidWord() is O(1) since we are given a dictionary.

for each element in the array:

T[i] = (IsValidWord(s.substring(i, 1)) && T[i+1])

|| (IsValidWord(s.substring(i, 2)) && T[i+2])

...

|| (IsValidWord(s.substring(i, n-i-1)) && T[n-1])

The result is T[0].

Complexity analysis

for each i, it takes O(n) to compute. and we start the computation from i = s.length - 1 and decrease i, and there are n element in the array to compute. So we get O(n^2) as the total time complexity.

Given two strings of a and b. Find the max window in a and b where the substrings in the window are anagrams.

Problem description:

1. Given two strings a and b, find the max value of "length" where for some i and j, a.substring(i, i+length) and b.substring(j, j+length) are anagrams.

2. Given two strings a and b, find the max window of matching patterns between a and b where the pattern matching doesn't have to be an exact match but also can be a permutation of each other. (which is essentially anagrams...)

Algorithm:

let m = a.length;
let n = b.length;
let matrix = int[m][n]

if a[i] == b[j] then matrix[i][j] = 1
else matrix[i][j] = 0

With the matrix populated by the routines above, find the largest square that meets the following two conditions:

• no column contains all 0s. For each column, at least one cell is 1
  
• no rows contains all 0s

The time complexity is O(n^2). I don't believe it can be done in O(nlogn). But I could be wrong.

Someone was asked the following question during interview: if m = n, would that help speeding up the complexity?

Well, I have no idea...

Checking whether a binary tree is balanced or not using recursion

Balanced: no path from root to leaf node is longer than another such path by 1.

The code is just so elegant...

this can be extended to general trees with more than two children by examining all children's maxDepth or minDepth.

ArrayList vs. LinkedList

There has been quite a bit discussion already. So just sum up by a few words:

ArrayList

• Lookup by index is O(1). (in Java, it implements RandomAccess) Suitable for dataset with random access pattern.
• Insertion can be expensive. Worst case O(n).
  1. Insertion to the middle can cause shifting.
  2. Insertion to the end can cause changing in the size of the internal storage, usually doubling.
• Deletion Depending on the implementation, it may vary a bit. Generally not bad.
  

LinkedList

• Lookup by index is O(n) in worst case. (in Java, it implements Queue). Suitable for data set that most of operations against it is iteration, or access mostly happens at the beginning or the end, such as FIFO queues, FILO stacks.
• Insertion is O(1)
• Deletion
  1. Deletion by reference is O(1)
  2. Deletion by value is as bad as lookup.

General thoughts:

Comparing two data structures, always at least think from two perspectives:

1. operations: lookup, insertion, deletion, manipulation.
2. input data characteristics: matching the characteristics with the cost of each operation to find the better data structure to solve your problem.

Hashtable

Disadvantage: generally, it wastes spaces.

Advantage: generally, it does fast lookup.

Three normal types: Chained, Chained Scattered, Open Scatter

http://tanksoftware.com/tutes/uni/hashing.html

A function that counts the number of 1s in an integer's binary expression

Finding all numbers which are smaller than K and only divisible by 2, 3, and 5

This may not be the most optimal solution...

Essentially we are looking for all possible {(x, y, z) | (2^x * 3^y * 5^z) <= K}

1. When x+y+z = 1, we get {2, 3, 5},
   
   
2. Let:
   
   S(i) = { (2^x * 3^y * 5^z) | (2^x * 3^y * 5^z) < K and (x + y + z = i)}
   
   S(i, j) = Union of all s in S(i-1, t) * j that:
   a. t and j belong to {2, 3, 5}
   b. t >= j
   c. s <= K
   
   for example:
   S(2,2) = {2, 3, 5} * 2 = {4, 6, 10}
   S(2,3) = {3, 5} * 3 = {9, 15}
   S(2,5) = {5} * 5 = {15}
   
   
3. S(i) = S(i-1, 2) + S(i -1, 3) + S(i-1, 5).
   
   for example:
   S(2) = S(2,2) + S(2,3) + S(2, 5)
   
   
4. Stops when S(i) is empty.